null hypothesis for single sample t

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The One-Sample t -Test

What is the one-sample t -test.

The one-sample t-test is a statistical hypothesis test used to determine whether an unknown population mean is different from a specific value.

When can I use the test?

You can use the test for continuous data. Your data should be a random sample from a normal population.

What if my data isn’t nearly normally distributed?

If your sample sizes are very small, you might not be able to test for normality. You might need to rely on your understanding of the data. When you cannot safely assume normality, you can perform a nonparametric test that doesn’t assume normality.

Using the one-sample t -test

See how to perform a one-sample t -test using statistical software.

• Download JMP to follow along using the sample data included with the software.
• To see more JMP tutorials, visit the JMP Learning Library .

The sections below discuss what we need for the test, checking our data, performing the test, understanding test results and statistical details.

What do we need?

For the one-sample t -test, we need one variable.

We also have an idea, or hypothesis, that the mean of the population has some value. Here are two examples:

• A hospital has a random sample of cholesterol measurements for men. These patients were seen for issues other than cholesterol. They were not taking any medications for high cholesterol. The hospital wants to know if the unknown mean cholesterol for patients is different from a goal level of 200 mg.
• We measure the grams of protein for a sample of energy bars. The label claims that the bars have 20 grams of protein. We want to know if the labels are correct or not.

One-sample t -test assumptions

For a valid test, we need data values that are:

• Independent (values are not related to one another).
• Continuous.
• Obtained via a simple random sample from the population.

Also, the population is assumed to be normally distributed .

One-sample t -test example

Imagine we have collected a random sample of 31 energy bars from a number of different stores to represent the population of energy bars available to the general consumer. The labels on the bars claim that each bar contains 20 grams of protein.

Table 1: Grams of protein in random sample of energy bars

If you look at the table above, you see that some bars have less than 20 grams of protein. Other bars have more. You might think that the data support the idea that the labels are correct. Others might disagree. The statistical test provides a sound method to make a decision, so that everyone makes the same decision on the same set of data values. 

Checking the data

Let’s start by answering: Is the t -test an appropriate method to test that the energy bars have 20 grams of protein ? The list below checks the requirements for the test.

• The data values are independent. The grams of protein in one energy bar do not depend on the grams in any other energy bar. An example of dependent values would be if you collected energy bars from a single production lot. A sample from a single lot is representative of that lot, not energy bars in general.
• The data values are grams of protein. The measurements are continuous.
• We assume the energy bars are a simple random sample from the population of energy bars available to the general consumer (i.e., a mix of lots of bars).
• We assume the population from which we are collecting our sample is normally distributed, and for large samples, we can check this assumption.

We decide that the t -test is an appropriate method.

Before jumping into analysis, we should take a quick look at the data. The figure below shows a histogram and summary statistics for the energy bars.

From a quick look at the histogram, we see that there are no unusual points, or outliers . The data look roughly bell-shaped, so our assumption of a normal distribution seems reasonable.

From a quick look at the statistics, we see that the average is 21.40, above 20. Does this  average from our sample of 31 bars invalidate the label's claim of 20 grams of protein for the unknown entire population mean? Or not?

How to perform the one-sample t -test

For the t -test calculations we need the mean, standard deviation and sample size. These are shown in the summary statistics section of Figure 1 above.

We round the statistics to two decimal places. Software will show more decimal places, and use them in calculations. (Note that Table 1 shows only two decimal places; the actual data used to calculate the summary statistics has more.)

We start by finding the difference between the sample mean and 20:

$ 21.40-20\ =\ 1.40$

Next, we calculate the standard error for the mean. The calculation is:

Standard Error for the mean = $ \frac{s}{\sqrt{n}}= \frac{2.54}{\sqrt{31}}=0.456 $

This matches the value in Figure 1 above.

We now have the pieces for our test statistic. We calculate our test statistic as:

$ t =  \frac{\text{Difference}}{\text{Standard Error}}= \frac{1.40}{0.456}=3.07 $

To make our decision, we compare the test statistic to a value from the t- distribution. This activity involves four steps.

• We calculate a test statistic. Our test statistic is 3.07.
• We decide on the risk we are willing to take for declaring a difference when there is not a difference. For the energy bar data, we decide that we are willing to take a 5% risk of saying that the unknown population mean is different from 20 when in fact it is not. In statistics-speak, we set α = 0.05. In practice, setting your risk level (α) should be made before collecting the data.

We find the value from the t- distribution based on our decision. For a t -test, we need the degrees of freedom to find this value. The degrees of freedom are based on the sample size. For the energy bar data:

degrees of freedom = $ n - 1 = 31 - 1 = 30 $

The critical value of t with α = 0.05 and 30 degrees of freedom is +/- 2.043. Most statistics books have look-up tables for the distribution. You can also find tables online. The most likely situation is that you will use software and will not use printed tables.

We compare the value of our statistic (3.07) to the t value. Since 3.07 > 2.043, we reject the null hypothesis that the mean grams of protein is equal to 20. We make a practical conclusion that the labels are incorrect, and the population mean grams of protein is greater than 20.

Statistical details

Let’s look at the energy bar data and the 1-sample t -test using statistical terms.

Our null hypothesis is that the underlying population mean is equal to 20. The null hypothesis is written as:

$ H_o:  \mathrm{\mu} = 20 $

The alternative hypothesis is that the underlying population mean is not equal to 20. The labels claiming 20 grams of protein would be incorrect. This is written as:

$ H_a:  \mathrm{\mu} ≠ 20 $

This is a two-sided test. We are testing if the population mean is different from 20 grams in either direction. If we can reject the null hypothesis that the mean is equal to 20 grams, then we make a practical conclusion that the labels for the bars are incorrect. If we cannot reject the null hypothesis, then we make a practical conclusion that the labels for the bars may be correct.

We calculate the average for the sample and then calculate the difference with the population mean, mu:

$  \overline{x} - \mathrm{\mu} $

We calculate the standard error as:

$ \frac{s}{ \sqrt{n}} $

The formula shows the sample standard deviation as s and the sample size as n .  

The test statistic uses the formula shown below:

$  \dfrac{\overline{x} - \mathrm{\mu}} {s / \sqrt{n}} $

We compare the test statistic to a t value with our chosen alpha value and the degrees of freedom for our data. Using the energy bar data as an example, we set α = 0.05. The degrees of freedom ( df ) are based on the sample size and are calculated as:

$ df = n - 1 = 31 - 1 = 30 $

Statisticians write the t value with α = 0.05 and 30 degrees of freedom as:

$ t_{0.05,30} $

The t value for a two-sided test with α = 0.05 and 30 degrees of freedom is +/- 2.042. There are two possible results from our comparison:

• The test statistic is less extreme than the critical  t  values; in other words, the test statistic is not less than -2.042, or is not greater than +2.042. You fail to reject the null hypothesis that the mean is equal to the specified value. In our example, you would be unable to conclude that the label for the protein bars should be changed.
• The test statistic is more extreme than the critical  t  values; in other words, the test statistic is less than -2.042, or is greater than +2.042. You reject the null hypothesis that the mean is equal to the specified value. In our example, you conclude that either the label should be updated or the production process should be improved to produce, on average, bars with 20 grams of protein.

Testing for normality

The normality assumption is more important for small sample sizes than for larger sample sizes.

Normal distributions are symmetric, which means they are “even” on both sides of the center. Normal distributions do not have extreme values, or outliers. You can check these two features of a normal distribution with graphs. Earlier, we decided that the energy bar data was “close enough” to normal to go ahead with the assumption of normality. The figure below shows a normal quantile plot for the data, and supports our decision.

You can also perform a formal test for normality using software. The figure below shows results of testing for normality with JMP software. We cannot reject the hypothesis of a normal distribution. 

We can go ahead with the assumption that the energy bar data is normally distributed.

What if my data are not from a Normal distribution?

If your sample size is very small, it is hard to test for normality. In this situation, you might need to use your understanding of the measurements. For example, for the energy bar data, the company knows that the underlying distribution of grams of protein is normally distributed. Even for a very small sample, the company would likely go ahead with the t -test and assume normality.

What if you know the underlying measurements are not normally distributed? Or what if your sample size is large and the test for normality is rejected? In this situation, you can use a nonparametric test. Nonparametric  analyses do not depend on an assumption that the data values are from a specific distribution. For the one-sample t ­-test, the one possible nonparametric test is the Wilcoxon Signed Rank test. 

Understanding p-values

Using a visual, you can check to see if your test statistic is more extreme than a specified value in the distribution. The figure below shows a t- distribution with 30 degrees of freedom.

Since our test is two-sided and we set α = 0.05, the figure shows that the value of 2.042 “cuts off” 5% of the data in the tails combined.

The next figure shows our results. You can see the test statistic falls above the specified critical value. It is far enough “out in the tail” to reject the hypothesis that the mean is equal to 20.

Putting it all together with Software

You are likely to use software to perform a t -test. The figure below shows results for the 1-sample t -test for the energy bar data from JMP software.  

The software shows the null hypothesis value of 20 and the average and standard deviation from the data. The test statistic is 3.07. This matches the calculations above.

The software shows results for a two-sided test and for one-sided tests. We want the two-sided test. Our null hypothesis is that the mean grams of protein is equal to 20. Our alternative hypothesis is that the mean grams of protein is not equal to 20.  The software shows a p- value of 0.0046 for the two-sided test. This p- value describes the likelihood of seeing a sample average as extreme as 21.4, or more extreme, when the underlying population mean is actually 20; in other words, the probability of observing a sample mean as different, or even more different from 20, than the mean we observed in our sample. A p -value of 0.0046 means there is about 46 chances out of 10,000. We feel confident in rejecting the null hypothesis that the population mean is equal to 20.

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One Sample T Test – Clearly Explained with Examples | ML+

• October 8, 2020
• Selva Prabhakaran

One sample T-Test tests if the given sample of observations could have been generated from a population with a specified mean.

If it is found from the test that the means are statistically different, we infer that the sample is unlikely to have come from the population.

For example: If you want to test a car manufacturer’s claim that their cars give a highway mileage of 20kmpl on an average. You sample 10 cars from the dealership, measure their mileage and use the T-test to determine if the manufacturer’s claim is true.

By end of this, you will know when and how to do the T-Test, the concept, math, how to set the null and alternate hypothesis, how to use the T-tables, how to understand the one-tailed and two-tailed T-Test and see how to implement in R and Python using a practical example.

Introduction

Purpose of one sample t test, how to set the null and alternate hypothesis, procedure to do one sample t test, one sample t test example, one sample t test implementation, how to decide which t test to perform two tailed, upper tailed or lower tailed.

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The ‘One sample T Test’ is one of the 3 types of T Tests . It is used when you want to test if the mean of the population from which the sample is drawn is of a hypothesized value. You will understand this statement better (and all of about One Sample T test) better by the end of this post.

T Test was first invented by William Sealy Gosset, in 1908. Since he used the pseudo name as ‘Student’ when publishing his method in the paper titled ‘Biometrika’, the test came to be know as Student’s T Test.

Since it assumes that the test statistic, typically the sample mean, follows the sampling distribution, the Student’s T Test is considered as a Parametric test.

The purpose of the One Sample T Test is to determine if a sample observations could have come from a process that follows a specific parameter (like the mean).

It is typically implemented on small samples.

For example, given a sample of 15 items, you want to test if the sample mean is the same as a hypothesized mean (population). That is, essentially you want to know if the sample came from the given population or not.

Let’s suppose, you want to test if the mean weight of a manufactured component (from a sample size 15) is of a particular value (55 grams), with a 99% confidence.

How did we determine One sample T-test is the right test for this?

Because, there is only one sample involved and you want to compare the mean of this sample against a particular (hypothesized) value..

To do this, you need to set up a null hypothesis and an alternate hypothesis .

The null hypothesis usually assumes that there is no difference in the sample means and the hypothesized mean (comparison mean). The purpose of the T Test is to test if the null hypothesis can be rejected or not.

Depending on the how the problem is stated, the alternate hypothesis can be one of the following 3 cases:

• Case 1: H1 : x̅ != µ. Used when the true sample mean is not equal to the comparison mean. Use Two Tailed T Test.
• Case 2: H1 : x̅ > µ. Used when the true sample mean is greater than the comparison mean. Use Upper Tailed T Test.
• Case 3: H1 : x̅ < µ. Used when the true sample mean is lesser than the comparison mean. Use Lower Tailed T Test.

Where x̅ is the sample mean and µ is the population mean for comparison. We will go more into the detail of these three cases after solving some practical examples.

Example 1: A customer service company wants to know if their support agents are performing on par with industry standards.

According to a report the standard mean resolution time is 20 minutes per ticket. The sample group has a mean at 21 minutes per ticket with a standard deviation of 7 minutes.

Can you tell if the company’s support performance is better than the industry standard or not?

Example 2: A farming company wants to know if a new fertilizer has improved crop yield or not.

Historic data shows the average yield of the farm is 20 tonne per acre. They decide to test a new organic fertilizer on a smaller sample of farms and observe the new yield is 20.175 tonne per acre with a standard deviation of 3.02 tonne for 12 different farms.

Did the new fertilizer work?

Step 1: Define the Null Hypothesis (H0) and Alternate Hypothesis (H1)

H0: Sample mean (x̅) = Hypothesized Population mean (µ)

H1: Sample mean (x̅) != Hypothesized Population mean (µ)

The alternate hypothesis can also state that the sample mean is greater than or less than the comparison mean.

Step 2: Compute the test statistic (T)

$$t = \frac{Z}{s} = \frac{\bar{X} – \mu}{\frac{\hat{\sigma}}{\sqrt{n}}}$$

where s is the standard error .

Step 3: Find the T-critical from the T-Table

Use the degree of freedom and the alpha level (0.05) to find the T-critical.

Step 4: Determine if the computed test statistic falls in the rejection region.

Alternately, simply compute the P-value. If it is less than the significance level (0.05 or 0.01), reject the null hypothesis.

Problem Statement:

We have the potato yield from 12 different farms. We know that the standard potato yield for the given variety is µ=20.

x = [21.5, 24.5, 18.5, 17.2, 14.5, 23.2, 22.1, 20.5, 19.4, 18.1, 24.1, 18.5]

Test if the potato yield from these farms is significantly better than the standard yield.

Step 1: Define the Null and Alternate Hypothesis

H0: x̅ = 20

H1: x̅ > 20

n = 12. Since this is one sample T test, the degree of freedom = n-1 = 12-1 = 11.

Let’s set alpha = 0.05, to meet 95% confidence level.

Step 2: Calculate the Test Statistic (T) 1. Calculate sample mean

$$\bar{X} = \frac{x_1 + x_2 + x_3 + . . + x_n}{n}$$

$$\bar{x} = 20.175$$

• Calculate sample standard deviation

$$\bar{\sigma} = \frac{(x_1 – \bar{x})^2 + (x_2 – \bar{x})^2 + (x_3 – \bar{x})^2 + . . + (x_n – \bar{x})^2}{n-1}$$

$$\sigma = 3.0211$$

• Substitute in the T Statistic formula

$$T = \frac{\bar{x} – \mu}{se} = \frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}}$$

$$T = (20.175 – 20)/(3.0211/\sqrt{12}) = 0.2006$$

Step 3: Find the T-Critical

Confidence level = 0.95, alpha=0.05. For one tailed test, look under 0.05 column. For d.o.f = 12 – 1 = 11, T-Critical = 1.796 .

Now you might wonder why ‘One Tailed test’ was chosen. This is because of the way you define the alternate hypothesis. Had the null hypothesis simply stated that the sample means is not equal to 20, then we would have gone for a two tailed test. More details about this topic in the next section.

Step 4: Does it fall in rejection region?

Since the computed T Statistic is less than the T-critical, it does not fall in the rejection region.

Clearly, the calculated T statistic does not fall in the rejection region. So, we do not reject the null hypothesis.

Since you want to perform a ‘One Tailed Greater than’ test (that is, the sample mean is greater than the comparison mean), you need to specify alternative='greater' in the t.test() function. Because, by default, the t.test() does a two tailed test (which is what you do when your alternate hypothesis simply states sample mean != comparison mean).

The P-value computed here is nothing but p = Pr(T > t) (upper-tailed), where t is the calculated T statistic.

In Python, One sample T Test is implemented in ttest_1samp() function in the scipy package. However, it does a Two tailed test by default , and reports a signed T statistic. That means, the reported P-value will always be computed for a Two-tailed test. To calculate the correct P value, you need to divide the output P-value by 2.

Apply the following logic if you are performing a one tailed test:

For greater than test: Reject H0 if p/2 < alpha (0.05). In this case, t will be greater than 0. For lesser than test: Reject H0 if p/2 < alpha (0.05). In this case, t will be less than 0.

Since it is one tailed test, the real p-value is 0.8446/2 = 0.4223. We do not rejecting the Null Hypothesis anyway.

The decision of whether the computed test statistic falls in the rejection region depends on how the alternate hypothesis is defined.

We know the Null Hypothesis is H0: µD = 0. Where, µD is the difference in the means, that is sample mean minus the comparison mean.

You can also write H0 as: x̅ = µ , where x̅ is sample mean and ‘µ’ is the comparison mean.

Case 1: If H1 : x̅ != µ , then rejection region lies on both tails of the T-Distribution (two-tailed). This means the alternate hypothesis just states the difference in means is not equal. There is no comparison if one of the means is greater or lesser than the other.

In this case, use Two Tailed T Test .

Here, P value = 2 . Pr(T > | t |)

Case 2: If H1: x̅ > µ , then rejection region lies on upper tail of the T-Distribution (upper-tailed). If the mean of the sample of interest is greater than the comparison mean. Example: If Component A has a longer time-to-failure than Component B.

In such case, use Upper Tailed based test.

Here, P-value = Pr(T > t)

Case 3: If H1: x̅ < µ , then rejection region lies on lower tail of the T-Distribution (lower-tailed). If the mean of the sample of interest is lesser than the comparison mean.

In such case, use lower tailed test.

Here, P-value = Pr(T < t)

Hope you are now familiar and clear about with the One Sample T Test. If some thing is still not clear, write in comment. Next, topic is Two sample T test . Stay tuned.

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Single Sample T-Test

The StatsTest Flow: Difference >> Continuous Variable of Interest >> One Sample Tests (single group) >> Normal Variable of Interest

Not sure this is the right statistical method? Use the Choose Your StatsTest workflow to select the right method.

What is a Single Sample T-Test?

The Single Sample T-Test is a statistical test used to determine if a single group is significantly different from a known or hypothesized population value on your variable of interest. Your variable of interest should be continuous and normally distributed and you should have enough data (more than 5 values).

The Single Sample T-Test is also called a One-Sample T-Test, Single Sample Student T-Test, or One-Sample Test of Means.

Assumptions for a Single Sample T-Test

Every statistical method has assumptions. Assumptions mean that your data must satisfy certain properties in order for statistical method results to be accurate.

The assumptions for the Single Sample T-Test include:

Normally Distributed

Random sample, enough data.

Let’s dive in to each one of these separately.

The variable that you care about (and want to see if it is different between your group and the population) must be continuous. Continuous means that the variable can take on any reasonable value.

Some good examples of continuous variables include age, weight, height, test scores, survey scores, yearly salary, etc.

If the variable that you care about is a proportion (48% of males voted vs 56% of females voted) and you have more than 5 in each group then you should use the One-Proportion Z-Test . If your variable of interest is a proportion and you have less than 5 in a group, you should use the Exact Test of Goodness of Fit .

Normally Distributed Variable of Interest

The variable that you care about must be spread out in a normal way. In statistics, this is called being normally distributed (aka it must look like a bell curve when you graph the data). Only use a single sample t-test with your data if the variable you care about is normally distributed.

If your variable is not normally distributed, you should use Single-Sample Wilcoxon Signed-Rank Test instead.

The data points for each group in your analysis must have come from a simple random sample. This means that if you wanted to see if drinking sugary soda makes you gain weight, you would need to randomly select a group of soda drinkers for your soda drinker group, and then you would compare that to a known population weight for non-sugary-soda drinkers.

The key here is that the data points for each group were randomly selected. This is important because if your group is not randomly determined then your analysis will be incorrect. In statistical terms this is called bias, or a tendency to have incorrect results because of bad data.

If you do not have a random sample, the conclusions you can draw from your results are very limited. You should try to get a simple random sample. If you have paired samples (2 measurements from the same group of subjects) then you should use a Paired Samples T-Test instead. If you want to compare 2 groups of subjects instead of a single group with a population mean, then you should use an Independent Samples T-Test instead

The sample size (or data set size) should be greater than 5 in your group. Some people argue for more than 15 or even 30, but more than 5 is probably sufficient.

It also depends on the expected size of the difference between groups. If you expect a large difference between groups, then you can get away with a smaller sample size. If you expect a small difference between groups, then you likely need a larger sample (30+).

If your sample size is greater than 30 (and you know the average and standard deviation or spread of the population values), you should run a Single Sample Z-Test instead.

When to use a Single Sample T-Test?

You should use a Single Sample T-Test in the following scenario:

• You want to know if one group is different from a known or hypothesized population value on your variable of interest
• Your variable of interest is continuous
• You have one group
• Your variable of interest is normally distributed

Let’s clarify these to help you know when to use a Single Sample T-Test.

You are looking for a statistical test to see whether a single group is significantly different from a population value on your variable of interest. This is a difference question. Other types of analyses include examining the relationship between two variables (correlation) or predicting one variable using another variable (prediction).

Continuous Data

Your variable of interest must be continuous. Continuous means that your variable of interest can basically take on any value, such as heart rate, height, weight, number of ice cream bars you can eat in 1 minute, etc.

Types of data that are NOT continuous include ordered data (such as finishing place in a race, best business rankings, etc.), categorical data (gender, eye color, race, etc.), or binary data (purchased the product or not, has the disease or not, etc.).

A Single Sample T-Test can only be used to compare a single group with a known population value on your variable of interest.

If you have three or more groups, you should use a One Way Anova analysis instead. If you have two groups to compare, you should use an Independent Samples T-Test instead.

Normally distributed was covered earlier and means that your variable of interest should look like a bell curve when you graph it as a histogram.

If you get a group of students to take a pre-test and the same students to take a post-test, you have two different variables for the same group of students, which would be paired data, in which case you would need to use a Paired Samples T-Test instead.

Single Sample T-Test Example

Group 1 : Received the experimental medical treatment. Population Value : On average in the population, it takes 12 days to recover from the disease Variable of interest : Time to recover from the disease in days.

In this example, group 1 is our treatment group because they received the experimental medical treatment. The population value is essentially our control group because they did not receive the treatment.

The null hypothesis, which is statistical lingo for what would happen if the treatment does nothing, is that group 1 and our population will recover from the disease in about the same number of days, on average. We are trying to determine if receiving the experimental medical treatment will shorten the number of days it takes for patients to recover from the disease.

As we run the experiment, we track how long it takes for each patient to fully recover from the disease. In order to use a Single Sample T-Test on our data, our variable of interest has to be normally distributed (bell curve shaped). In this case, recovery from the disease in days is normal for our treatment group.

After the experiment is over, we compare our treatment group to the population value on our variable of interest (days to fully recover) using a Single Sample T-Test. When we run the analysis, we get a t-statistic and a p-value. The t-statistic is a measure of how different our group is from the population value on our recovery variable of interest. A p-value is the chance of seeing our results assuming the treatment actually doesn’t do anything. A p-value less than or equal to 0.05 means that our result is statistically significant and we can trust that the difference is not due to chance alone.

Frequently Asked Questions

Q: What is the difference between a single sample t-test and a one sample t-test? A: Nothing. They are two names for the same analysis.

Q: What if I don’t know the population average for my variable of interest? A: You cannot run a single sample t-test without a comparison group or value. You either need to collect data for a control group or find data on what the population average is.

Q: How do I run a single sample t-test in SPSS, R, SAS, or STATA? A: This resource is focused on helping you pick the right statistical method every time. There are many resources available to help you figure out how to run this method with your data: SPSS article: https://libguides.library.kent.edu/SPSS/OneSampletTest SPSS video: https://www.youtube.com/watch?v=2zVeV1ohGCU R article: http://www.sthda.com/english/wiki/one-sample-t-test-in-r R video: https://www.youtube.com/watch?v=kvmSAXhX9Hs

If you still can’t figure something out, feel free to reach out .

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One Sample t Test

Spss tutorials: one sample t test.

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• rank transform converts a set of data values by ordering them from smallest to largest, and then assigning a rank to each value. In SPSS, the Rank Cases procedure can be used to compute the rank transform of a variable." href="https://libguides.library.kent.edu/SPSS/RankCases" style="" >Computing Variables: Rank Transforms (Rank Cases)
• Weighting Cases
• Sorting Data
• Grouping Data
• Descriptive Stats for One Numeric Variable (Explore)
• Descriptive Stats for One Numeric Variable (Frequencies)
• Descriptive Stats for Many Numeric Variables (Descriptives)
• Descriptive Stats by Group (Compare Means)
• Frequency Tables
• Working with "Check All That Apply" Survey Data (Multiple Response Sets)
• Chi-Square Test of Independence
• Pearson Correlation
• Paired Samples t Test
• Independent Samples t Test
• One-Way ANOVA
• How to Cite the Tutorials

Sample Data Files

Our tutorials reference a dataset called "sample" in many examples. If you'd like to download the sample dataset to work through the examples, choose one of the files below:

• Data definitions (*.pdf)
• Data - Comma delimited (*.csv)
• Data - Tab delimited (*.txt)
• Data - Excel format (*.xlsx)
• Data - SAS format (*.sas7bdat)
• Data - SPSS format (*.sav)
• SPSS Syntax (*.sps) Syntax to add variable labels, value labels, set variable types, and compute several recoded variables used in later tutorials.
• SAS Syntax (*.sas) Syntax to read the CSV-format sample data and set variable labels and formats/value labels.

The One Sample t Test examines whether the mean of a population is statistically different from a known or hypothesized value. The One Sample t Test is a parametric test.

This test is also known as:

• Single Sample t Test

The variable used in this test is known as:

• Test variable

In a One Sample t Test, the test variable's mean is compared against a "test value", which is a known or hypothesized value of the mean in the population. Test values may come from a literature review, a trusted research organization, legal requirements, or industry standards. For example:

• A particular factory's machines are supposed to fill bottles with 150 milliliters of product. A plant manager wants to test a random sample of bottles to ensure that the machines are not under- or over-filling the bottles.
• The United States Environmental Protection Agency (EPA) sets clearance levels for the amount of lead present in homes: no more than 10 micrograms per square foot on floors and no more than 100 micrograms per square foot on window sills ( as of December 2020 ). An inspector wants to test if samples taken from units in an apartment building exceed the clearance level.

Common Uses

The One Sample  t  Test is commonly used to test the following:

• Statistical difference between a mean and a known or hypothesized value of the mean in the population.
• This approach involves creating a change score from two variables, and then comparing the mean change score to zero, which will indicate whether any change occurred between the two time points for the original measures. If the mean change score is not significantly different from zero, no significant change occurred.

Note: The One Sample t Test can only compare a single sample mean to a specified constant. It can not compare sample means between two or more groups. If you wish to compare the means of multiple groups to each other, you will likely want to run an Independent Samples t Test (to compare the means of two groups) or a One-Way ANOVA (to compare the means of two or more groups).

Data Requirements

Your data must meet the following requirements:

• Test variable that is continuous (i.e., interval or ratio level)
• There is no relationship between scores on the test variable
• Violation of this assumption will yield an inaccurate p value
• Random sample of data from the population
• Non-normal population distributions, especially those that are thick-tailed or heavily skewed, considerably reduce the power of the test
• Among moderate or large samples, a violation of normality may still yield accurate p values
• Homogeneity of variances (i.e., variances approximately equal in both the sample and population)
• No outliers

The null hypothesis ( H 0 ) and (two-tailed) alternative hypothesis ( H 1 ) of the one sample T test can be expressed as:

H 0 : µ =  µ 0   ("the population mean is equal to the [proposed] population mean") H 1 : µ ≠  µ 0   ("the population mean is not equal to the [proposed] population mean")

where µ is the "true" population mean and µ 0 is the proposed value of the population mean.

Test Statistic

The test statistic for a One Sample t Test is denoted t , which is calculated using the following formula:

$$ t = \frac{\overline{x}-\mu{}_{0}}{s_{\overline{x}}} $$

$$ s_{\overline{x}} = \frac{s}{\sqrt{n}} $$

\(\mu_{0}\) = The test value -- the proposed constant for the population mean \(\bar{x}\) = Sample mean \(n\) = Sample size (i.e., number of observations) \(s\) = Sample standard deviation \(s_{\bar{x}}\) = Estimated standard error of the mean ( s /sqrt( n ))

The calculated t value is then compared to the critical t value from the t distribution table with degrees of freedom df = n - 1 and chosen confidence level. If the calculated t value > critical t value, then we reject the null hypothesis.

Data Set-Up

Your data should include one continuous, numeric variable (represented in a column) that will be used in the analysis. The variable's measurement level should be defined as Scale in the Variable View window.

Run a One Sample t Test

To run a One Sample t Test in SPSS, click  Analyze > Compare Means > One-Sample T Test .

The One-Sample T Test window opens where you will specify the variables to be used in the analysis. All of the variables in your dataset appear in the list on the left side. Move variables to the Test Variable(s) area by selecting them in the list and clicking the arrow button.

A Test Variable(s): The variable whose mean will be compared to the hypothesized population mean (i.e., Test Value). You may run multiple One Sample t Tests simultaneously by selecting more than one test variable. Each variable will be compared to the same Test Value. 

B Test Value: The hypothesized population mean against which your test variable(s) will be compared.

C Estimate effect sizes: Optional. If checked, will print effect size statistics -- namely, Cohen's d -- for the test(s). (Note: Effect sizes calculations for t tests were first added to SPSS Statistics in version 27, making them a relatively recent addition. If you do not see this option when you use SPSS, check what version of SPSS you're using.)

D Options: Clicking Options will open a window where you can specify the Confidence Interval Percentage and how the analysis will address Missing Values (i.e., Exclude cases analysis by analysis or Exclude cases listwise ). Click Continue when you are finished making specifications.

Click OK to run the One Sample t Test.

Problem Statement

According to the CDC , the mean height of U.S. adults ages 20 and older is about 66.5 inches (69.3 inches for males, 63.8 inches for females).

In our sample data, we have a sample of 435 college students from a single college. Let's test if the mean height of students at this college is significantly different than 66.5 inches using a one-sample t test. The null and alternative hypotheses of this test will be:

H 0 : µ Height = 66.5  ("the mean height is equal to 66.5") H 1 : µ Height ≠ 66.5  ("the mean height is not equal to 66.5")

Before the Test

In the sample data, we will use the variable Height , which a continuous variable representing each respondent’s height in inches. The heights exhibit a range of values from 55.00 to 88.41 ( Analyze > Descriptive Statistics > Descriptives ).

Let's create a histogram of the data to get an idea of the distribution, and to see if  our hypothesized mean is near our sample mean. Click Graphs > Legacy Dialogs > Histogram . Move variable Height to the Variable box, then click OK .

To add vertical reference lines at the mean (or another location), double-click on the plot to open the Chart Editor, then click Options > X Axis Reference Line . In the Properties window, you can enter a specific location on the x-axis for the vertical line, or you can choose to have the reference line at the mean or median of the sample data (using the sample data). Click Apply to make sure your new line is added to the chart. Here, we have added two reference lines: one at the sample mean (the solid black line), and the other at 66.5 (the dashed red line).

From the histogram, we can see that height is relatively symmetrically distributed about the mean, though there is a slightly longer right tail. The reference lines indicate that sample mean is slightly greater than the hypothesized mean, but not by a huge amount. It's possible that our test result could come back significant.

Running the Test

To run the One Sample t Test, click  Analyze > Compare Means > One-Sample T Test.  Move the variable Height to the Test Variable(s) area. In the Test Value field, enter 66.5.

If you are using SPSS Statistics 27 or later :

If you are using SPSS Statistics 26 or earlier :

Two sections (boxes) appear in the output: One-Sample Statistics and One-Sample Test . The first section, One-Sample Statistics , provides basic information about the selected variable, Height , including the valid (nonmissing) sample size ( n ), mean, standard deviation, and standard error. In this example, the mean height of the sample is 68.03 inches, which is based on 408 nonmissing observations.

The second section, One-Sample Test , displays the results most relevant to the One Sample t Test. 

A Test Value : The number we entered as the test value in the One-Sample T Test window.

B t Statistic : The test statistic of the one-sample t test, denoted t . In this example, t = 5.810. Note that t is calculated by dividing the mean difference (E) by the standard error mean (from the One-Sample Statistics box).

C df : The degrees of freedom for the test. For a one-sample t test, df = n - 1; so here, df = 408 - 1 = 407.

D Significance (One-Sided p and Two-Sided p): The p-values corresponding to one of the possible one-sided alternative hypotheses (in this case, µ Height > 66.5) and two-sided alternative hypothesis (µ Height ≠ 66.5), respectively. In our problem statement above, we were only interested in the two-sided alternative hypothesis.

E Mean Difference : The difference between the "observed" sample mean (from the One Sample Statistics box) and the "expected" mean (the specified test value (A)). The sign of the mean difference corresponds to the sign of the t value (B). The positive t value in this example indicates that the mean height of the sample is greater than the hypothesized value (66.5).

F Confidence Interval for the Difference : The confidence interval for the difference between the specified test value and the sample mean.

Decision and Conclusions

Recall that our hypothesized population value was 66.5 inches, the [approximate] average height of the overall adult population in the U.S. Since p < 0.001, we reject the null hypothesis that the mean height of students at this college is equal to the hypothesized population mean of 66.5 inches and conclude that the mean height is significantly different than 66.5 inches.

Based on the results, we can state the following:

• There is a significant difference in the mean height of the students at this college and the overall adult population in the U.S. ( p < .001).
• The average height of students at this college is about 1.5 inches taller than the U.S. adult population average (95% CI [1.013, 2.050]).
• << Previous: Pearson Correlation
• Next: Paired Samples t Test >>
• Last Updated: May 10, 2024 1:32 PM
• URL: https://libguides.library.kent.edu/SPSS

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4.1: One-Sample t-Test

• Last updated
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• Page ID 1734

• John H. McDonald
• University of Delaware

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Learning Objectives

• Use Student's \(t\)–test for one sample when you have one measurement variable and a theoretical expectation of what the mean should be under the null hypothesis. It tests whether the mean of the measurement variable is different from the null expectation.

There are several statistical tests that use the \(t\)-distribution and can be called a \(t\) - test. One is Student's \(t\) - test for one sample, named after "Student," the pseudonym that William Gosset used to hide his employment by the Guinness brewery in the early 1900s (they had a rule that their employees weren't allowed to publish, and Guinness didn't want other employees to know that they were making an exception for Gosset). Student's \(t\) - test for one sample compares a sample to a theoretical mean. It has so few uses in biology that I didn't cover it in previous editions of this Handbook, but then I recently found myself using it (McDonald and Dunn 2013), so here it is.

When to use it

Use Student's \(t\)-test when you have one measurement variable, and you want to compare the mean value of the measurement variable to some theoretical expectation. It is commonly used in fields such as physics (you've made several observations of the mass of a new subatomic particle—does the mean fit the mass predicted by the Standard Model of particle physics?) and product testing (you've measured the amount of drug in several aliquots from a new batch—is the mean of the new batch significantly less than the standard you've established for that drug?). It's rare to have this kind of theoretical expectation in biology, so you'll probably never use the one-sample \(t\)-test.

I've had a hard time finding a real biological example of a one-sample \(t\)-test, so imagine that you're studying joint position sense, our ability to know what position our joints are in without looking or touching. You want to know whether people over- or underestimate their knee angle. You blindfold \(10\) volunteers, bend their knee to a \(120^{\circ}\) angle for a few seconds, then return the knee to a \(90^{\circ}\) angle. Then you ask each person to bend their knee to the \(120^{\circ}\) angle. The measurement variable is the angle of the knee, and the theoretical expectation from the null hypothesis is \(120^{\circ}\). You get the following imaginary data:

If the null hypothesis were true that people don't over- or underestimate their knee angle, the mean of these \(10\) numbers would be \(120\). The mean of these ten numbers is \(117.2\); the one-sample \(t\)–test will tell you whether that is significantly different from \(120\).

Null hypothesis

The statistical null hypothesis is that the mean of the measurement variable is equal to a number that you decided on before doing the experiment. For the knee example, the biological null hypothesis is that people don't under- or overestimate their knee angle. You decided to move people's knees to \(120^{\circ}\), so the statistical null hypothesis is that the mean angle of the subjects' knees will be \(120^{\circ}\).

How the test works

Calculate the test statistic,\(t_s\), using this formula:

\[t_s=\frac{(\bar{x}-\mu _\theta )}{(s/\sqrt{n})}\]

where \(\bar{x}\) is the sample mean, \(\mu\) is the mean expected under the null hypothesis, \(s\) is the sample standard deviation and \(n\) is the sample size. The test statistic, \(t_s\), gets bigger as the difference between the observed and expected means gets bigger, as the standard deviation gets smaller, or as the sample size gets bigger.

Applying this formula to the imaginary knee position data gives a \(t\)-value of \(-3.69\).

You calculate the probability of getting the observed \(t_s\) value under the null hypothesis using the t-distribution. The shape of the \(t\)-distribution, and thus the probability of getting a particular \(t_s\) value, depends on the number of degrees of freedom. The degrees of freedom for a one-sample \(t\)-test is the total number of observations in the group minus \(1\). For our example data, the \(P\) value for a \(t\)-value of \(-3.69\) with \(9\) degrees of freedom is \(0.005\), so you would reject the null hypothesis and conclude that people return their knee to a significantly smaller angle than the original position.

Assumptions

The \(t\) - test assumes that the observations within each group are normally distributed. If the distribution is symmetrical, such as a flat or bimodal distribution, the one-sample \(t\) - test is not at all sensitive to the non-normality; you will get accurate estimates of the \(P\) value, even with small sample sizes. A severely skewed distribution can give you too many false positives unless the sample size is large (above \(50\) or so). If your data are severely skewed and you have a small sample size, you should try a data transformation to make them less skewed. With large sample sizes (simulations I've done suggest \(50\) is large enough), the one-sample \(t\) - test will give accurate results even with severely skewed data.

McDonald and Dunn (2013) measured the correlation of transferrin (labeled red) and Rab-10 (labeled green) in five cells. The biological null hypothesis is that transferrin and Rab-10 are not colocalized (found in the same subcellular structures), so the statistical null hypothesis is that the correlation coefficient between red and green signals in each cell image has a mean of zero. The correlation coefficients were \(0.52,\; 0.20,\; 0.59,\; 0.62\) and \(0.60\) in the five cells. The mean is \(0.51\), which is highly significantly different from \(0\) (\(t=6.46,\; 4d.f.,\; P=0.003\)), indicating that transferrin and Rab-10 are colocalized in these cells.

Graphing the results

Because you're just comparing one observed mean to one expected value, you probably won't put the results of a one-sample \(t\) - test in a graph. If you've done a bunch of them, I guess you could draw a bar graph with one bar for each mean, and a dotted horizontal line for the null expectation.

Similar tests

The paired t –test is a special case of the one-sample \(t\) - test; it tests the null hypothesis that the mean difference between two measurements (such as the strength of the right arm minus the strength of the left arm) is equal to zero. Experiments that use a paired t –test are much more common in biology than experiments using the one-sample \(t\) - test, so I treat the paired \(t\)-test as a completely different test.

The two-sample t –test compares the means of two different samples. If one of your samples is very large, you may be tempted to treat the mean of the large sample as a theoretical expectation, but this is incorrect. For example, let's say you want to know whether college softball pitchers have greater shoulder flexion angles than normal people. You might be tempted to look up the "normal" shoulder flexion angle (\(150^{\circ}\)) and compare your data on pitchers to the normal angle using a one-sample \(t\) - test. However, the "normal" value doesn't come from some theory, it is based on data that has a mean, a standard deviation, and a sample size, and at the very least you should dig out the original study and compare your sample to the sample the \(150^{\circ}\) "normal" was based on, using a two-sample \(t\)-test that takes the variation and sample size of both samples into account.

How to do the test

Spreadsheets.

I have set up a spreadsheet to perform the one-sample \(t\)–test onesamplettest.xls. It will handle up to \(1000\) observations.

There are web pages to do the one-sample \(t\)–test here and here .

Salvatore Mangiafico's \(R\) Companion has a sample R program for the one-sample t –test .

You can use PROC TTEST for Student's \(t\)-test; the CLASS parameter is the nominal variable, and the VAR parameter is the measurement variable. Here is an example program for the joint position sense data above. Note that \(H0\) parameter for the theoretical value is \(H\) followed by the numeral zero, not a capital letter \(O\).

DATA jps; INPUT angle; DATALINES; 120.6 116.4 117.2 118.1 114.1 116.9 113.3 121.1 116.9 117.0 ; PROC TTEST DATA=jps H0=50; VAR angle; RUN;

The output includes some descriptive statistics, plus the \(t\)-value and \(P\) value. For these data, the \(P\) value is \(0.005\).

DF t Value Pr > |t| 9 -3.69 0.0050

Power analysis

To estimate the sample size you to detect a significant difference between a mean and a theoretical value, you need the following:

• the effect size, or the difference between the observed mean and the theoretical value that you hope to detect
• the standard deviation
• alpha, or the significance level (usually \(0.05\))
• beta, the probability of accepting the null hypothesis when it is false (\(0.50,\; 0.80\) and \(0.90\) are common values)

The G*Power program will calculate the sample size needed for a one-sample \(t\)-test. Choose "t tests" from the "Test family" menu and "Means: Difference from constant (one sample case)" from the "Statistical test" menu. Click on the "Determine" button and enter the theoretical value ("Mean \(H0\)") and a mean with the smallest difference from the theoretical that you hope to detect ("Mean \(H1\)"). Enter an estimate of the standard deviation. Click on "Calculate and transfer to main window". Change "tails" to two, set your alpha (this will almost always be \(0.05\)) and your power (\(0.5,\; 0.8,\; or\; 0.9\) are commonly used).

As an example, let's say you want to follow up the knee joint position sense study that I made up above with a study of hip joint position sense. You're going to set the hip angle to \(70^{\circ}\) (Mean \(H0=70\)) and you want to detect an over- or underestimation of this angle of \(1^{\circ}\), so you set Mean \(H1=71\). You don't have any hip angle data, so you use the standard deviation from your knee study and enter \(2.4\) for SD. You want to do a two-tailed test at the \(P<0.05\) level, with a probability of detecting a difference this large, if it exists, of \(90\%\) (\(1-\text {beta}=0.90\)). Entering all these numbers in G*Power gives a sample size of \(63\) people.

• McDonald, J.H., and K.W. Dunn. 2013. Statistical tests for measures of colocalization in biological microscopy. Journal of Microscopy 252: 295-302.

t-test Calculator

Table of contents

Welcome to our t-test calculator! Here you can not only easily perform one-sample t-tests , but also two-sample t-tests , as well as paired t-tests .

Do you prefer to find the p-value from t-test, or would you rather find the t-test critical values? Well, this t-test calculator can do both! 😊

What does a t-test tell you? Take a look at the text below, where we explain what actually gets tested when various types of t-tests are performed. Also, we explain when to use t-tests (in particular, whether to use the z-test vs. t-test) and what assumptions your data should satisfy for the results of a t-test to be valid. If you've ever wanted to know how to do a t-test by hand, we provide the necessary t-test formula, as well as tell you how to determine the number of degrees of freedom in a t-test.

When to use a t-test?

A t-test is one of the most popular statistical tests for location , i.e., it deals with the population(s) mean value(s).

There are different types of t-tests that you can perform:

• A one-sample t-test;
• A two-sample t-test; and
• A paired t-test.

In the next section , we explain when to use which. Remember that a t-test can only be used for one or two groups . If you need to compare three (or more) means, use the analysis of variance ( ANOVA ) method.

The t-test is a parametric test, meaning that your data has to fulfill some assumptions :

• The data points are independent; AND
• The data, at least approximately, follow a normal distribution .

If your sample doesn't fit these assumptions, you can resort to nonparametric alternatives. Visit our Mann–Whitney U test calculator or the Wilcoxon rank-sum test calculator to learn more. Other possibilities include the Wilcoxon signed-rank test or the sign test.

Which t-test?

Your choice of t-test depends on whether you are studying one group or two groups:

One sample t-test

Choose the one-sample t-test to check if the mean of a population is equal to some pre-set hypothesized value .

The average volume of a drink sold in 0.33 l cans — is it really equal to 330 ml?

The average weight of people from a specific city — is it different from the national average?

Two-sample t-test

Choose the two-sample t-test to check if the difference between the means of two populations is equal to some pre-determined value when the two samples have been chosen independently of each other.

In particular, you can use this test to check whether the two groups are different from one another .

The average difference in weight gain in two groups of people: one group was on a high-carb diet and the other on a high-fat diet.

The average difference in the results of a math test from students at two different universities.

This test is sometimes referred to as an independent samples t-test , or an unpaired samples t-test .

Paired t-test

A paired t-test is used to investigate the change in the mean of a population before and after some experimental intervention , based on a paired sample, i.e., when each subject has been measured twice: before and after treatment.

In particular, you can use this test to check whether, on average, the treatment has had any effect on the population .

The change in student test performance before and after taking a course.

The change in blood pressure in patients before and after administering some drug.

How to do a t-test?

So, you've decided which t-test to perform. These next steps will tell you how to calculate the p-value from t-test or its critical values, and then which decision to make about the null hypothesis.

Decide on the alternative hypothesis :

Use a two-tailed t-test if you only care whether the population's mean (or, in the case of two populations, the difference between the populations' means) agrees or disagrees with the pre-set value.

Use a one-tailed t-test if you want to test whether this mean (or difference in means) is greater/less than the pre-set value.

Compute your T-score value :

Formulas for the test statistic in t-tests include the sample size , as well as its mean and standard deviation . The exact formula depends on the t-test type — check the sections dedicated to each particular test for more details.

Determine the degrees of freedom for the t-test:

The degrees of freedom are the number of observations in a sample that are free to vary as we estimate statistical parameters. In the simplest case, the number of degrees of freedom equals your sample size minus the number of parameters you need to estimate . Again, the exact formula depends on the t-test you want to perform — check the sections below for details.

The degrees of freedom are essential, as they determine the distribution followed by your T-score (under the null hypothesis). If there are d degrees of freedom, then the distribution of the test statistics is the t-Student distribution with d degrees of freedom . This distribution has a shape similar to N(0,1) (bell-shaped and symmetric) but has heavier tails . If the number of degrees of freedom is large (>30), which generically happens for large samples, the t-Student distribution is practically indistinguishable from N(0,1).

💡 The t-Student distribution owes its name to William Sealy Gosset, who, in 1908, published his paper on the t-test under the pseudonym "Student". Gosset worked at the famous Guinness Brewery in Dublin, Ireland, and devised the t-test as an economical way to monitor the quality of beer. Cheers! 🍺🍺🍺

p-value from t-test

Recall that the p-value is the probability (calculated under the assumption that the null hypothesis is true) that the test statistic will produce values at least as extreme as the T-score produced for your sample . As probabilities correspond to areas under the density function, p-value from t-test can be nicely illustrated with the help of the following pictures:

The following formulae say how to calculate p-value from t-test. By cdf t,d we denote the cumulative distribution function of the t-Student distribution with d degrees of freedom:

p-value from left-tailed t-test:

p-value = cdf t,d (t score )

p-value from right-tailed t-test:

p-value = 1 − cdf t,d (t score )

p-value from two-tailed t-test:

p-value = 2 × cdf t,d (−|t score |)

or, equivalently: p-value = 2 − 2 × cdf t,d (|t score |)

However, the cdf of the t-distribution is given by a somewhat complicated formula. To find the p-value by hand, you would need to resort to statistical tables, where approximate cdf values are collected, or to specialized statistical software. Fortunately, our t-test calculator determines the p-value from t-test for you in the blink of an eye!

t-test critical values

Recall, that in the critical values approach to hypothesis testing, you need to set a significance level, α, before computing the critical values , which in turn give rise to critical regions (a.k.a. rejection regions).

Formulas for critical values employ the quantile function of t-distribution, i.e., the inverse of the cdf :

Critical value for left-tailed t-test: cdf t,d -1 (α)

critical region:

(-∞, cdf t,d -1 (α)]

Critical value for right-tailed t-test: cdf t,d -1 (1-α)

[cdf t,d -1 (1-α), ∞)

Critical values for two-tailed t-test: ±cdf t,d -1 (1-α/2)

(-∞, -cdf t,d -1 (1-α/2)] ∪ [cdf t,d -1 (1-α/2), ∞)

To decide the fate of the null hypothesis, just check if your T-score lies within the critical region:

If your T-score belongs to the critical region , reject the null hypothesis and accept the alternative hypothesis.

If your T-score is outside the critical region , then you don't have enough evidence to reject the null hypothesis.

How to use our t-test calculator

Choose the type of t-test you wish to perform:

A one-sample t-test (to test the mean of a single group against a hypothesized mean);

A two-sample t-test (to compare the means for two groups); or

A paired t-test (to check how the mean from the same group changes after some intervention).

Two-tailed;

Left-tailed; or

Right-tailed.

This t-test calculator allows you to use either the p-value approach or the critical regions approach to hypothesis testing!

Enter your T-score and the number of degrees of freedom . If you don't know them, provide some data about your sample(s): sample size, mean, and standard deviation, and our t-test calculator will compute the T-score and degrees of freedom for you .

Once all the parameters are present, the p-value, or critical region, will immediately appear underneath the t-test calculator, along with an interpretation!

One-sample t-test

The null hypothesis is that the population mean is equal to some value μ 0 \mu_0 μ 0 ​ .

The alternative hypothesis is that the population mean is:

• different from μ 0 \mu_0 μ 0 ​ ;
• smaller than μ 0 \mu_0 μ 0 ​ ; or
• greater than μ 0 \mu_0 μ 0 ​ .

One-sample t-test formula :

• μ 0 \mu_0 μ 0 ​ — Mean postulated in the null hypothesis;
• n n n — Sample size;
• x ˉ \bar{x} x ˉ — Sample mean; and
• s s s — Sample standard deviation.

Number of degrees of freedom in t-test (one-sample) = n − 1 n-1 n − 1 .

The null hypothesis is that the actual difference between these groups' means, μ 1 \mu_1 μ 1 ​ , and μ 2 \mu_2 μ 2 ​ , is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the difference μ 1 − μ 2 \mu_1 - \mu_2 μ 1 ​ − μ 2 ​ is:

• Different from Δ \Delta Δ ;
• Smaller than Δ \Delta Δ ; or
• Greater than Δ \Delta Δ .

In particular, if this pre-determined difference is zero ( Δ = 0 \Delta = 0 Δ = 0 ):

The null hypothesis is that the population means are equal.

The alternate hypothesis is that the population means are:

• μ 1 \mu_1 μ 1 ​ and μ 2 \mu_2 μ 2 ​ are different from one another;
• μ 1 \mu_1 μ 1 ​ is smaller than μ 2 \mu_2 μ 2 ​ ; and
• μ 1 \mu_1 μ 1 ​ is greater than μ 2 \mu_2 μ 2 ​ .

Formally, to perform a t-test, we should additionally assume that the variances of the two populations are equal (this assumption is called the homogeneity of variance ).

There is a version of a t-test that can be applied without the assumption of homogeneity of variance: it is called a Welch's t-test . For your convenience, we describe both versions.

Two-sample t-test if variances are equal

Use this test if you know that the two populations' variances are the same (or very similar).

Two-sample t-test formula (with equal variances) :

where s p s_p s p ​ is the so-called pooled standard deviation , which we compute as:

• Δ \Delta Δ — Mean difference postulated in the null hypothesis;
• n 1 n_1 n 1 ​ — First sample size;
• x ˉ 1 \bar{x}_1 x ˉ 1 ​ — Mean for the first sample;
• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• n 2 n_2 n 2 ​ — Second sample size;
• x ˉ 2 \bar{x}_2 x ˉ 2 ​ — Mean for the second sample; and
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

Number of degrees of freedom in t-test (two samples, equal variances) = n 1 + n 2 − 2 n_1 + n_2 - 2 n 1 ​ + n 2 ​ − 2 .

Two-sample t-test if variances are unequal (Welch's t-test)

Use this test if the variances of your populations are different.

Two-sample Welch's t-test formula if variances are unequal:

• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

The number of degrees of freedom in a Welch's t-test (two-sample t-test with unequal variances) is very difficult to count. We can approximate it with the help of the following Satterthwaite formula :

Alternatively, you can take the smaller of n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 as a conservative estimate for the number of degrees of freedom.

🔎 The Satterthwaite formula for the degrees of freedom can be rewritten as a scaled weighted harmonic mean of the degrees of freedom of the respective samples: n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 , and the weights are proportional to the standard deviations of the corresponding samples.

As we commonly perform a paired t-test when we have data about the same subjects measured twice (before and after some treatment), let us adopt the convention of referring to the samples as the pre-group and post-group.

The null hypothesis is that the true difference between the means of pre- and post-populations is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the actual difference between these means is:

Typically, this pre-determined difference is zero. We can then reformulate the hypotheses as follows:

The null hypothesis is that the pre- and post-means are the same, i.e., the treatment has no impact on the population .

The alternative hypothesis:

• The pre- and post-means are different from one another (treatment has some effect);
• The pre-mean is smaller than the post-mean (treatment increases the result); or
• The pre-mean is greater than the post-mean (treatment decreases the result).

Paired t-test formula

In fact, a paired t-test is technically the same as a one-sample t-test! Let us see why it is so. Let x 1 , . . . , x n x_1, ... , x_n x 1 ​ , ... , x n ​ be the pre observations and y 1 , . . . , y n y_1, ... , y_n y 1 ​ , ... , y n ​ the respective post observations. That is, x i , y i x_i, y_i x i ​ , y i ​ are the before and after measurements of the i -th subject.

For each subject, compute the difference, d i : = x i − y i d_i := x_i - y_i d i ​ := x i ​ − y i ​ . All that happens next is just a one-sample t-test performed on the sample of differences d 1 , . . . , d n d_1, ... , d_n d 1 ​ , ... , d n ​ . Take a look at the formula for the T-score :

Δ \Delta Δ — Mean difference postulated in the null hypothesis;

n n n — Size of the sample of differences, i.e., the number of pairs;

x ˉ \bar{x} x ˉ — Mean of the sample of differences; and

s s s  — Standard deviation of the sample of differences.

Number of degrees of freedom in t-test (paired): n − 1 n - 1 n − 1

t-test vs Z-test

We use a Z-test when we want to test the population mean of a normally distributed dataset, which has a known population variance . If the number of degrees of freedom is large, then the t-Student distribution is very close to N(0,1).

Hence, if there are many data points (at least 30), you may swap a t-test for a Z-test, and the results will be almost identical. However, for small samples with unknown variance, remember to use the t-test because, in such cases, the t-Student distribution differs significantly from the N(0,1)!

🙋 Have you concluded you need to perform the z-test? Head straight to our z-test calculator !

What is a t-test?

A t-test is a widely used statistical test that analyzes the means of one or two groups of data. For instance, a t-test is performed on medical data to determine whether a new drug really helps.

What are different types of t-tests?

Different types of t-tests are:

• One-sample t-test;
• Two-sample t-test; and
• Paired t-test.

How to find the t value in a one sample t-test?

To find the t-value:

• Subtract the null hypothesis mean from the sample mean value.
• Divide the difference by the standard deviation of the sample.
• Multiply the resultant with the square root of the sample size.

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ol{padding-top:0px;}.css-4okk7a ul:not(:first-child),.css-4okk7a ol:not(:first-child){padding-top:4px;} Test setup

Choose test type

t-test for the population mean, μ, based on one independent sample . Null hypothesis H 0 : μ = μ 0  

Alternative hypothesis H 1

Test details

Significance level α

The probability that we reject a true H 0 (type I error).

Degrees of freedom

Calculated as sample size minus one.

Test results

• Calculators
• Descriptive Statistics
• Merchandise
• Which Statistics Test?

Single Sample T-Test Calculator

A single sample t-test (or one sample t-test) is used to compare the mean of a single sample of scores to a known or hypothetical population mean. So, for example, it could be used to determine whether the mean diastolic blood pressure of a particular group differs from 85, a value determined by a previous study.

Requirements

• The data is normally distributed
• Scale of measurement should be interval or ratio
• A randomized sample from a defined population

Null Hypothesis

H 0 : M - μ = 0, where M is the sample mean and μ is the population or hypothesized mean.

As above, the null hypothesis is that there is no difference between the sample mean and the known or hypothesized population mean.

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• One sample t test

A one sample t test compares the mean with a hypothetical value. In most cases, the hypothetical value comes from theory. For example, if you express your data as 'percent of control', you can test whether the average differs significantly from 100. The hypothetical value can also come from previous data. For example, compare whether the mean systolic blood pressure differs from 135, a value determined in a previous study.

1. Choose data entry format

Caution: Changing format will erase your data.

2. Specify the hypothetical mean value

3. enter data, 4. view the results, learn more about the one sample t test.

In this article you will learn the requirements and assumptions of a one sample t test, how to format and interpret the results of a one sample t test, and when to use different types of t tests.

One sample t test: Overview

The one sample t test, also referred to as a single sample t test, is a statistical hypothesis test used to determine whether the mean calculated from sample data collected from a single group is different from a designated value specified by the researcher. This designated value does not come from the data itself, but is an external value chosen for scientific reasons. Often, this designated value is a mean previously established in a population, a standard value of interest, or a mean concluded from other studies. Like all hypothesis testing, the one sample t test determines if there is enough evidence reject the null hypothesis (H0) in favor of an alternative hypothesis (H1). The null hypothesis for a one sample t test can be stated as: "The population mean equals the specified mean value." The alternative hypothesis for a one sample t test can be stated as: "The population mean is different from the specified mean value."

The one sample t test differs from most statistical hypothesis tests because it does not compare two separate groups or look at a relationship between two variables. It is a straightforward comparison between data gathered on a single variable from one population and a specified value defined by the researcher. The one sample t test can be used to look for a difference in only one direction from the standard value (a one-tailed t test ) or can be used to look for a difference in either direction from the standard value (a two-tailed t test ).

Requirements and Assumptions for a one sample t test

A one sample t test should be used only when data has been collected on one variable for a single population and there is no comparison being made between groups. For a valid one sample t test analysis, data values must be all of the following:

The one sample t test assumes that all "errors" in the data are independent. The term "error" refers to the difference between each value and the group mean. The results of a t test only make sense when the scatter is random - that whatever factor caused a value to be too high or too low affects only that one value. Prism cannot test this assumption, but there are graphical ways to explore data to verify this assumption is met.

A t test is only appropriate to apply in situations where data represent variables that are continuous measurements. As they rely on the calculation of a mean value, variables that are categorical should not be analyzed using a t test.

The results of a t test should be based on a random sample and only be generalized to the larger population from which samples were drawn.

As with all parametric hypothesis testing, the one sample t test assumes that you have sampled your data from a population that follows a normal (or Gaussian) distribution. While this assumption is not as important with large samples, it is important with small sample sizes, especially less than 10. If your data do not come from a Gaussian distribution , there are three options to accommodate this. One option is to transform the values to make the distribution more Gaussian, perhaps by transforming all values to their reciprocals or logarithms. Another choice is to use the Wilcoxon signed rank nonparametric test instead of the t test. A final option is to use the t test anyway, knowing that the t test is fairly robust to departures from a Gaussian distribution with large samples.

How to format a one sample t test

Ideally, data for a one sample t test should be collected and entered as a single column from which a mean value can be easily calculated. If data is entered on a table with multiple subcolumns, Prism requires one of the following choices to be selected to perform the analysis:

• Each subcolumn of data can be analyzed separately
• An average of the values in the columns across each row can be calculated, and the analysis conducted on this new stack of means, or
• All values in all columns can be treated as one sample of data (paying no attention to which row or column any values are in).

How the one sample t test calculator works

Prism calculates the t ratio by dividing the difference between the actual and hypothetical means by the standard error of the actual mean. The equation is written as follows, where x is the calculated mean, μ is the hypothetical mean (specified value), S is the standard deviation of the sample, and n is the sample size:

A p value is computed based on the calculated t ratio and the numbers of degrees of freedom present (which equals sample size minus 1). The one sample t test calculator assumes it is a two-tailed one sample t test, meaning you are testing for a difference in either direction from the specified value.

How to interpret results of a one sample t test

As discussed, a one sample t test compares the mean of a single column of numbers against a hypothetical mean. This hypothetical mean can be based upon a specific standard or other external prediction. The test produces a P value which requires careful interpretation.

The p value answers this question: If the data were sampled from a Gaussian population with a mean equal to the hypothetical value you entered, what is the chance of randomly selecting N data points and finding a mean as far (or further) from the hypothetical value as observed here?

If the p value is large (usually defined to mean greater than 0.05), the data do not give you any reason to conclude that the population mean differs from the designated value to which it has been compared. This is not the same as saying that the true mean equals the hypothetical value, but rather states that there is no evidence of a difference. Thus, we cannot reject the null hypothesis (H0).

If the p value is small (usually defined to mean less than or equal to 0.05), then it is unlikely that the discrepancy observed between the sample mean and hypothetical mean is due to a coincidence arising from random sampling. There is evidence to reject the idea that the difference is coincidental and conclude instead that the population has a mean that is different from the hypothetical value to which it has been compared. The difference is statistically significant, and the null hypothesis is therefore rejected.

If the null hypothesis is rejected, the question of whether the difference is scientifically important still remains. The confidence interval can be a useful tool in answering this question. Prism reports the 95% confidence interval for the difference between the actual and hypothetical mean. In interpreting these results, one can be 95% sure that this range includes the true difference. It requires scientific judgment to determine if this difference is truly meaningful.

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When to use different types of t tests

There are three types of t tests which can be used for hypothesis testing:

• Independent two-sample (or unpaired) t test
• Paired sample t test

As described, a one sample t test should be used only when data has been collected on one variable for a single population and there is no comparison being made between groups. It only applies when the mean value for data is intended to be compared to a fixed and defined number.

In most cases involving data analysis, however, there are multiple groups of data either representing different populations being compared, or the same population being compared at different times or conditions. For these situations, it is not appropriate to use a one sample t test. Other types of t tests are appropriate for these specific circumstances:

Independent Two-Sample t test (Unpaired t test)

The independent sample t test, also referred to as the unpaired t test, is used to compare the means of two different samples. The independent two-sample t test comes in two different forms:

• the standard Student's t test, which assumes that the variance of the two groups are equal.
• the Welch's t test , which is less restrictive compared to the original Student's test. This is the test where you do not assume that the variance is the same in the two groups, which results in fractional degrees of freedom.

The two methods give very similar results when the sample sizes are equal and the variances are similar.

Paired Sample t test

The paired sample t test is used to compare the means of two related groups of samples. Put into other words, it is used in a situation where you have two values (i.e., a pair of values) for the same group of samples. Often these two values are measured from the same samples either at two different times, under two different conditions, or after a specific intervention.

You can perform multiple independent two-sample comparison tests simultaneously in Prism. Select from parametric and nonparametric tests and specify if the data are unpaired or paired. Try performing a t test with a 30-day free trial of Prism .

Watch this video to learn how to choose between a paired and unpaired t test.

Example of how to apply the appropriate t test

"Alkaline" labeled bottled drinking water has become fashionable over the past several years. Imagine we have collected a random sample of 30 bottles of "alkaline" drinking water from a number of different stores to represent the population of "alkaline" bottled water for a particular brand available to the general consumer. The labels on each of the bottles claim that the pH of the "alkaline" water is 8.5. A laboratory then proceeds to measure the exact pH of the water in each bottle.

Table 1: pH of water in random sample of "alkaline bottled water"

If you look at the table above, you see that some bottles have a pH measured to be lower than 8.5, while other bottles have a pH measured to be higher. What can the data tell us about the actual pH levels found in this brand of "alkaline" water bottles marketed to the public as having a pH of 8.5? Statistical hypothesis testing provides a sound method to evaluate this question. Which specific test to use, however, depends on the specific question being asked.

Is a t test appropriate to apply to this data?

Let's start by asking: Is a t test an appropriate method to analyze this set of pH data? The following list reviews the requirements and assumptions for using a t test:

• Independent sampling : In an independent sample t test, the data values are independent. The pH of one bottle of water does not depend on the pH of any other water bottle. (An example of dependent values would be if you collected water bottles from a single production lot. A sample from a single lot is representative only of that lot, not of alkaline bottled water in general).
• Continuous variable : The data values are pH levels, which are numerical measurements that are continuous.
• Random sample : We assume the water bottles are a simple random sample from the population of "alkaline" water bottles produced by this brand as they are a mix of many production lots.
• Normal distribution : We assume the population from which we collected our samples has pH levels that are normally distributed. To verify this, we should visualize the data graphically. The figure below shows a histogram for the pH measurements of the water bottles. From a quick look at the histogram, we see that there are no unusual points, or outliers. The data look roughly bell-shaped, so our assumption of a normal distribution seems reasonable. The QQ plot can also be used to graphically assess normality and is the preferred choice when the sample size is small.

Based upon these features and assumptions being met, we can conclude that a t test is an appropriate method to be applied to this set of data.

Which t test is appropriate to use?

The next decision is which t test to apply, and this depends on the exact question we would like our analysis to answer. This example illustrates how each type of t test could be chosen for a specific analysis, and why the one sample t test is the correct choice to determine if the measured pH of the bottled water samples match the advertised pH of 8.5.

We could be interested in determining whether a certain characteristic of a water bottle is associated with having a higher or lower pH, such as whether bottles are glass or plastic. For this questions, we would effectively be dividing the bottles into 2 separate groups and comparing the means of the pH between the 2 groups. For this analysis, we would elect to use a two sample t test because we are comparing the means of two independent groups.

We could also be interested in learning if pH is affected by a water bottle being opened and exposed to the air for a week. In this case, each original sample would be tested for pH level after a week had elapsed and the water had been exposed to the air, creating a second set of sample data. To evaluate whether this exposure affected pH, we would again be comparing two different groups of data, but this time the data are in paired samples each having an original pH measurement and a second measurement from after the week of exposure to the open air. For this analysis, it is appropriate to use a paired t test so that data for each bottle is assembled in rows, and the change in pH is considered bottle by bottle.

Returning to the original question we set out to answer-whether bottled water that is advertised to have a pH of 8.5 actually meets this claim-it is now clear that neither an independent two sample t test or a paired t test would be appropriate. In this case, all 30 pH measurements are sampled from one group representing bottled drinking water labeled "alkaline" available to the general consumer. We wish to compare this measured mean with an expected advertised value of 8.5. This is the exact situation for which one should employ a one sample t test!

From a quick look at the descriptive statistics, we see that the mean of the sample measurements is 8.513, slightly above 8.5. Does this average from our sample of 30 bottles validate the advertised claim of pH 8.5? By applying Prism's one sample t test analysis to this data set, we will get results by which we can evaluate whether the null hypothesis (that there is no difference between the mean pH level in the water bottles and the pH level advertised on the bottles) should be accepted or rejected.

How to Perform a One Sample T Test in Prism

In prior versions of Prism, the one sample t test and the Wilcoxon rank sum tests were computed as part of Prism's Column Statistics analysis. Now, starting with Prism 8, performing one sample t tests is even easier with a separate analysis in Prism.

Steps to perform a one sample t test in Prism

• Create a Column data table.
• Enter each data set in a single Y column so all values from each group are stacked into a column. Prism will perform a one sample t test (or Wilcoxon rank sum test) on each column you enter.
• Click Analyze, look in the list of Column analyses, and choose one sample t test and Wilcoxon test.

It's that simple! Prism streamlines your t test analysis so you can make more accurate and more informed data interpretations. Start your 30-day free trial of Prism and try performing your first one sample t test in Prism.

Watch this video for a step-by-step tutorial on how to perform a t test in Prism.

We Recommend:

Analyze, graph and present your scientific work easily with GraphPad Prism. No coding required.

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Single Sample t Test

Menu location: Analysis_Parametric_Single Sample t .

This function gives a single sample Student t test with a confidence interval for the mean difference.

The single sample t method tests a null hypothesis that the population mean is equal to a specified value. If this value is zero (or not entered) then the confidence interval for the sample mean is given ( Altman, 1991; Armitage and Berry, 1994 ).

The test statistic is calculated as:

- where x bar is the sample mean, s² is the sample variance, n is the sample size, µ is the specified population mean and t is a Student t quantile with n-1 degrees of freedom.

Power is calculated as the power achieved with the given sample size and variance for detecting the observed mean difference with a two-sided type I error probability of (100-CI%)% ( Dupont, 1990 ).

Test workbook (Parametric worksheet: Systolic BP).

Consider 20 first year resident female doctors drawn at random from one area, resting systolic blood pressures measured using an electronic sphygmomanometer were:

From previous large studies of women drawn at random from the healthy general public, a resting systolic blood pressure of 120 mm Hg was predicted as the population mean for the relevant age group. To analyse these data in StatsDirect first prepare a workbook column containing the 20 data above or open the test workbook and select the single sample t test from the parametric methods section of the analysis menu. Select the column marked "Systolic BP" when prompted and enter the population mean as 120.

For this example:

Single sample t test

Sample name: Systolic BP

Sample mean = 130.05

Population mean = 120

Sample size n = 20

Sample sd = 9.960316

95% confidence interval for mean difference = 5.388429 to 14.711571

t = 4.512404

One sided P = .0001

Two sided P = .0002

Power (for 5% significance) = 98.71%

A null hypothesis of no difference between sample and population means has clearly been rejected. Using the 95% CI we expect the mean systolic BP for this population of doctors to be at least 5 mm Hg greater than the age and sex matched general public, lying somewhere between 125 and 135 mm Hg.

confidence intervals

Copyright © 2000-2024 StatsDirect Limited, all rights reserved. Download here .

One-Sample T-test in R

• Discussion (1)
• Printable version

What is one-sample t-test?

Research questions and statistical hypotheses, formula of one-sample t-test, install ggpubr r package for data visualization, r function to compute one-sample t-test, import your data into r, check your data, visualize your data using box plots, preleminary test to check one-sample t-test assumptions, compute one-sample t-test, interpretation of the result, access to the values returned by t.test() function, online one-sample t-test calculator.

Generally, the theoretical mean comes from:

• a previous experiment. For example, compare whether the mean weight of mice differs from 200 mg, a value determined in a previous study.
• or from an experiment where you have control and treatment conditions. If you express your data as “percent of control”, you can test whether the average value of treatment condition differs significantly from 100.

Note that, one-sample t-test can be used only, when the data are normally distributed . This can be checked using Shapiro-Wilk test .

Related Book:

Typical research questions are:

• whether the mean ( \(m\) ) of the sample is equal to the theoretical mean ( \(\mu\) )?
• whether the mean ( \(m\) ) of the sample is less than the theoretical mean ( \(\mu\) )?
• whether the mean ( \(m\) ) of the sample is greater than the theoretical mean ( \(\mu\) )?

In statistics, we can define the corresponding null hypothesis ( \(H_0\) ) as follow:

• \(H_0: m = \mu\)
• \(H_0: m \leq \mu\)
• \(H_0: m \geq \mu\)

The corresponding alternative hypotheses ( \(H_a\) ) are as follow:

• \(H_a: m \ne \mu\) (different)
• \(H_a: m > \mu\) (greater)
• \(H_a: m (less)
• Hypotheses 1) are called two-tailed tests
• Hypotheses 2) and 3) are called one-tailed tests

The t-statistic can be calculated as follow:

\[ t = \frac{m-\mu}{s/\sqrt{n}} \]

• m is the sample mean
• n is the sample size
• s is the sample standard deviation with \(n-1\) degrees of freedom
• \(\mu\) is the theoretical value

We can compute the p-value corresponding to the absolute value of the t-test statistics (|t|) for the degrees of freedom (df): \(df = n - 1\) .

How to interpret the results?

If the p-value is inferior or equal to the significance level 0.05, we can reject the null hypothesis and accept the alternative hypothesis. In other words, we conclude that the sample mean is significantly different from the theoretical mean.

Visualize your data and compute one-sample t-test in R

You can draw R base graps as described at this link: R base graphs . Here, we’ll use the ggpubr R package for an easy ggplot2-based data visualization

• Install the latest version from GitHub as follow (recommended):
• Or, install from CRAN as follow:

To perform one-sample t-test, the R function t.test () can be used as follow:

• x : a numeric vector containing your data values
• mu : the theoretical mean. Default is 0 but you can change it.
• alternative : the alternative hypothesis. Allowed value is one of “two.sided” (default), “greater” or “less”.

Prepare your data as specified here: Best practices for preparing your data set for R

Save your data in an external .txt tab or .csv files

Import your data into R as follow:

Here, we’ll use an example data set containing the weight of 10 mice.

We want to know, if the average weight of the mice differs from 25g?

• Min. : the minimum value
• 1st Qu. : The first quartile. 25% of values are lower than this.
• Median : the median value. Half the values are lower; half are higher.
• 3rd Qu. : the third quartile. 75% of values are higher than this.
• Max. : the maximum value

One-Sample Student’s T-test in R

• Is this a large sample ? - No, because n < 30.
• Since the sample size is not large enough (less than 30, central limit theorem), we need to check whether the data follow a normal distribution .

How to check the normality?

Read this article: Normality Test in R .

Briefly, it’s possible to use the Shapiro-Wilk normality test and to look at the normality plot .

• Null hypothesis: the data are normally distributed
• Alternative hypothesis: the data are not normally distributed

From the output, the p-value is greater than the significance level 0.05 implying that the distribution of the data are not significantly different from normal distribtion. In other words, we can assume the normality.

• Visual inspection of the data normality using Q-Q plots (quantile-quantile plots). Q-Q plot draws the correlation between a given sample and the normal distribution.

From the normality plots, we conclude that the data may come from normal distributions.

Note that, if the data are not normally distributed, it’s recommended to use the non parametric one-sample Wilcoxon rank test.

We want to know, if the average weight of the mice differs from 25g (two-tailed test)?

In the result above :

• t is the t-test statistic value (t = -9.078),
• df is the degrees of freedom (df= 9),
• p-value is the significance level of the t-test (p-value = 7.95310^{-6}).
• conf.int is the confidence interval of the mean at 95% (conf.int = [17.8172, 20.6828]);
• sample estimates is he mean value of the sample (mean = 19.25).
• if you want to test whether the mean weight of mice is less than 25g (one-tailed test), type this:
• Or, if you want to test whether the mean weight of mice is greater than 25g (one-tailed test), type this:

The p-value of the test is 7.95310^{-6}, which is less than the significance level alpha = 0.05. We can conclude that the mean weight of the mice is significantly different from 25g with a p-value = 7.95310^{-6}.

The result of t.test() function is a list containing the following components:

• statistic : the value of the t test statistics
• parameter : the degrees of freedom for the t test statistics
• p.value : the p-value for the test
• conf.int : a confidence interval for the mean appropriate to the specified alternative hypothesis .
• estimate : the means of the two groups being compared (in the case of independent t test ) or difference in means (in the case of paired t test ).

The format of the R code to use for getting these values is as follow:

You can perform one-sample t-test , online , without any installation by clicking the following link:

One-sample wilcoxon test (non-parametric)

This analysis has been performed using R software (ver. 3.2.4).

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One Sample t-test Calculator

p-value (one-tailed) = 0.1245

p-value (two-tailed) = 0.3232

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Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

2 Replies to “One Sample t-test Calculator”

This seems like a great tool!

This is an excellent help in my course of studies. Thanks a million

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